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Let $G'$ be the commutator subgroup of a group $G$ and $G^*=\langle g^{-1}\alpha(g)\mid g\in G, \alpha\in Aut(G)\rangle$.

We know that always $G'\leq G^*$.

It is clear that if $Inn(G)=Aut(G)$, then $G'=G^*$.

Also if $G$ is a non abelian simple group or perfect group, then $G'=G^*=G$.

Now do there exist a group such that $Inn(G)\neq Aut(G)$ and $G'=G^*\neq G$?

Thank you

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3 Answers 3

Take $G = D_{5} = \langle \sigma, \tau \mid \sigma^5 = \tau^2 = 1, \tau \sigma \tau^{-1} = \sigma^{-1} \rangle$, the dihedral group of order $10$. Clearly, $G' = G^* = \langle \sigma \rangle$. Note that the automorphism $\sigma^i\tau^j \mapsto \sigma^{2i}\tau^j$ is not an inner automorphism.

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FWIW this is the smallest example, and is part of a whole family of examples including dihedrals of order 4k+2 for any k ≥ 2. k=1 only fails because Aut=Inn. –  Jack Schmidt May 13 '13 at 21:10

Serkan's answer is part of a more general family and a more general idea.

The more general idea is to include non-inner automorphisms that create no new subgroup fusion. The easiest way to do this is with power automorphisms, and the simplest examples of those are automorphisms that raise a single generator to a power.

The more general family is parameterized by pairs $(n,d)$ where $n$ is a positive integer and $1 \neq d \neq n-1$, but $d$ divides $n-1$. Let $A=\operatorname{AGL}(1,\newcommand{\znz}{\mathbb{Z}/n\mathbb{Z}}\znz)$ consist of the affine transformations of the line over $\znz$. The derived subgroup $D=A'$ consists of the translations. Choose some element $\zeta$ of multiplicative order $d$ mod $n$, and let $G$ be the subgroup generated by multiplication by $\zeta$ and by the translations. Explicitly: $$\begin{array}{rcl} A &=& \left\{ \begin{bmatrix} \alpha & \beta \\ 0 & 1 \end{bmatrix} : \alpha \in \znz^\times, \beta \in \znz \right\} \\ D &=& \left\{ \begin{bmatrix} 1 & \beta \\ 0 & 1 \end{bmatrix} : \beta \in \znz \right\} \\ G &=& \left\{ \begin{bmatrix} \zeta^i & \beta \\ 0 & 1 \end{bmatrix} : 0 \leq i < d, \beta \in \znz \right\} \\ \end{array}$$

For example, take $n=5$ and $d=2$, to get $D$ is a cyclic group of order 5, $G$ is dihedral of order 10, and $A$ is isomorphic to the normalizer of a Sylow 5-subgroup ($D$) in $S_5$.

Notice that $Z(G)=1$ since $d\neq 1$, so $G \cong \operatorname{Inn}(G)$. We'll show $A \cong \operatorname{Aut}(G)$ so that $D=G'=G^*$ as requested. Since $d \neq n-1$, $A \neq G$.

Notice that $D=G'$ (not just $D=A'$) so $D$ is characteristic in $G$, so automorphisms of $G$ restrict to automorphisms of the cyclic group $D$. Let $f$ be an automorphism of $G$, and choose $\beta, \bar\alpha, \bar\beta \in \znz$ so that $$\begin{array}{rcl} f\left( \begin{bmatrix} 1 & 1 \\0 & 1 \end{bmatrix} \right) &=& \begin{bmatrix} 1 & \beta \\ 0 & 1 \end{bmatrix} \\ f\left( \begin{bmatrix} \zeta & 0 \\0 & 1 \end{bmatrix} \right) &=& \begin{bmatrix} \bar\alpha & \bar\beta \\ 0 & 1 \end{bmatrix} \end{array}$$ Notice that $f$ is determined by these numbers since $G$ is generated by those two matrices, and that $\beta, \bar\alpha \in \znz^\times$. Now $$ \begin{bmatrix} \zeta & 0 \\0 & 1 \end{bmatrix} \cdot \begin{bmatrix} 1 & 1 \\0 & 1 \end{bmatrix} \cdot {\begin{bmatrix} \zeta & 0 \\0 & 1 \end{bmatrix}}^{-1} = \begin{bmatrix} 1 & \zeta \\0 & 1 \end{bmatrix} $$ so applying $f$ we get $$ \begin{bmatrix} 1 & \bar\alpha\cdot\beta \\0 & 1 \end{bmatrix} = \begin{bmatrix} \bar\alpha & \bar\beta \\0 & 1 \end{bmatrix} \cdot \begin{bmatrix} 1 & \beta \\0 & 1 \end{bmatrix} \cdot {\begin{bmatrix} \bar\alpha & \bar\beta \\0 & 1 \end{bmatrix}}^{-1} = \begin{bmatrix} 1 & \zeta\cdot\beta \\0 & 1 \end{bmatrix} $$ hence $\bar\alpha \beta = \zeta\beta$ and since $\beta \in \znz^\times$ (lest $D \cap \ker(f) \neq 1$), we get $\bar\alpha=\zeta$. Hence every automorphism of $G$ is conjugation by an element of $A$, namely $$\bar f = \begin{bmatrix} 1/\beta & \bar\beta/(\beta(\zeta-1)) \\ 0 & 1 \end{bmatrix} \in A. $$

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Maybe I should explicitly say, $D = G' \leq G^* = [G,A] \leq [A,A] = D$, so that we do have $G' = G^*$. –  Jack Schmidt May 13 '13 at 21:26

Take $G = S_6$, the symmetric group of degree $6$. It is known that $G' = A_6$ is the subgroup consisting of all even permutations. It is also not difficult to see that when $a \in \mathrm{Aut}(G)$ and $g \in G$, permutations $g$ and $a(g)$ must have the same sign (because $\mathrm{Aut}(G)$ acts trivially on $G/G' \simeq \mathbb{Z} / 2 \mathbb{Z}$), which means that $g^{-1}a(g) \in G'$. So, $G^* = G' \neq G$. But it is known that $\mathrm{Inn}(G) \subsetneq \mathrm{Aut}(G)$, so we have the desired example.

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